Integration — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for integration.

UT-1: Integration by Parts — Cyclic Integral Trap

Question:

(a) Evaluate $\displaystyle\int e^{2x}\sin x\,dx$.

(b) A student sets up $I = \int e^{2x}\sin x\,dx$Applies integration by parts twice, and gets $I = \text{(something)} - I$Then concludes $I = 0$. Explain the error.

[Difficulty: hard. Tests cyclic integration by parts and the algebraic step of collecting the $I$ terms.]

Solution:

(a) Let $I = \displaystyle\int e^{2x}\sin x\,dx$.

First application: u = e^{2x}$$dv = \sin x\,dx. Then du = 2e^{2x}\,dx$$v = -\cos x.

$I = -e^{2x}\cos x + \int 2e^{2x}\cos x\,dx$

Second application on $\int e^{2x}\cos x\,dx$: $u = e^{2x}$, $dv = \cos x\,dx$. Then $du = 2e^{2x}\,dx$, $v = \sin x$.

$\int e^{2x}\cos x\,dx = e^{2x}\sin x - \int 2e^{2x}\sin x\,dx = e^{2x}\sin x - I$

Substitute back:

$I = -e^{2x}\cos x + 2e^{2x}\sin x - 2I$

$3I = e^{2x}(2\sin x - \cos x)$

$I = \frac{e^{2x}(2\sin x - \cos x)}{3} + C$

(b) The student”s error is that when they got $I = \text{(something)} - I$They incorrectly concluded $I = 0$. The correct step is to add $I$ to both sides to get $2I = \text{(something)}$Then divide by $2$ (or in this case $3$). The cyclic nature of the integral means $I$ appears on both sides, but this does not mean $I = 0$ — it means $I$ can be solved for algebraically.

UT-2: Partial Fractions with Irreducible Quadratic

Question:

(a) Express $\dfrac{3x - 1}{(x + 1)(x^2 + 1)}$ in partial fractions.

(b) Hence evaluate $\displaystyle\int \frac{3x - 1}{(x + 1)(x^2 + 1)}\,dx$.

(c) A student attempts to split the denominator as $\dfrac{A}{x + 1} + \dfrac{B}{x^2 + 1}$ and gets $A = 3$, $B = -1$. Show that this is incorrect.

[Difficulty: hard. Tests partial fraction decomposition with an irreducible quadratic factor and the correct form of the decomposition.]

Solution:

(a) Since $x^2 + 1$ is irreducible (discriminant $= -4 \lt 0$), the correct form is:

$\frac{3x - 1}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}$

$3x - 1 = A(x^2 + 1) + (Bx + C)(x + 1)$

$3x - 1 = Ax^2 + A + Bx^2 + Bx + Cx + C$

$3x - 1 = (A + B)x^2 + (B + C)x + (A + C)$

Comparing coefficients:

• $x^2$: $A + B = 0$ \quad (i)
• $x$: $B + C = 3$ \quad (ii)
• constant: $A + C = -1$ \quad (iii)

From (i): $B = -A$. From (iii): $C = -1 - A$.

Substitute into (ii): $-A + (-1 - A) = 3 \implies -2A - 1 = 3 \implies -2A = 4 \implies A = -2$.

Then $B = 2$ and $C = -1 - (-2) = 1$.

$\frac{3x - 1}{(x + 1)(x^2 + 1)} = \frac{-2}{x + 1} + \frac{2x + 1}{x^2 + 1}$

(b)

$\int \frac{3x - 1}{(x + 1)(x^2 + 1)}\,dx = \int \frac{-2}{x + 1}\,dx + \int \frac{2x + 1}{x^2 + 1}\,dx$

$= -2\ln\lvert x + 1 \rvert + \int \frac{2x}{x^2 + 1}\,dx + \int \frac{1}{x^2 + 1}\,dx$

$= -2\ln\lvert x + 1 \rvert + \ln(x^2 + 1) + \arctan x + C$

(c) If the student uses $\dfrac{A}{x + 1} + \dfrac{B}{x^2 + 1}$Then:

$3x - 1 = A(x^2 + 1) + B(x + 1) = Ax^2 + A + Bx + B$

Comparing coefficients: $A = 0$ (from $x^2$ term), $B = 3$ (from $x$ term), and $A + B = 3 = -1$ (from constant term), which is a contradiction ($3 \neq -1$). This shows the form $\dfrac{B}{x^2 + 1}$ is incorrect — the numerator must be linear: $\dfrac{Bx + C}{x^2 + 1}$.

UT-3: Improper Integral — Convergence with Parameter

Question:

(a) Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^p}\,dx$ converges or diverges, stating the condition on $p$.

(b) Evaluate $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$ and explain why it converges despite the integrand being unbounded.

(c) A student claims that since $\dfrac{1}{\sqrt{x}} \to \infty$ as $x \to 0^+$The integral must diverge. Explain the error.

[Difficulty: hard. Tests improper integrals at both endpoints and the $p$-test for convergence.]

Solution:

(a)

$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b \to \infty}\int_1^b x^{-p}\,dx$

Case $p \neq 1$:

$= \lim_{b \to \infty}\left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty}\left(\frac{b^{1-p}}{1-p} - \frac{1}{1-p}\right)$

• If $p \gt 1$: $1 - p \lt 0$So $b^{1-p} \to 0$ and the integral converges to $\dfrac{1}{p - 1}$.
• If $p \lt 1$: $1 - p \gt 0$So $b^{1-p} \to \infty$ and the integral diverges.

Case $p = 1$:

$\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty}\ln b = \infty \quad \text{(diverges)}$

The integral converges if and only if $p \gt 1$.

(b)

$\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{a \to 0^+}\int_a^1 x^{-1/2}\,dx = \lim_{a \to 0^+}\left[2\sqrt{x}\right]_a^1 = \lim_{a \to 0^+}(2 - 2\sqrt{a}) = 2$

The integral converges to $2$. Although the integrand is unbounded at $x = 0$The area under the curve is finite because the singularity is integrable (the exponent $-\frac{1}{2} \gt -1$).

(c) The student confuses the behaviour of the integrand with the behaviour of the integral. An unbounded integrand does not necessarily produce a divergent integral. The key question is whether the area accumulates to a finite value. For $\frac{1}{\sqrt{x}}$ near $x = 0$The function grows, but slowly enough that the total area remains bounded. The $p$-test shows that $\int_0^1 x^{-p}\,dx$ converges when $p \lt 1$.

Integration Tests

Tests synthesis of integration with other topics.

IT-1: Area Between Curves with Trigonometry

Question:

Find the total area enclosed by the curves $y = \sin x$ and $y = \cos x$ for $0 \le x \le \pi$.

[Difficulty: hard. Combines curve sketching, intersection finding, and definite integration with absolute value.]

Solution:

Find intersection points in $[0, \pi]$:

$\sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}$

In $[0, \pi]$This is the only intersection.

Check which curve is on top:

• For $0 \lt x \lt \frac{\pi}{4}$: $\cos x \gt \sin x$ (e.g., at $x = 0$: $\cos 0 = 1 \gt 0 = \sin 0$).
• For $\frac{\pi}{4} \lt x \lt \pi$: $\sin x \gt \cos x$ (e.g., at $x = \frac{\pi}{2}$: $\sin\frac{\pi}{2} = 1 \gt 0 = \cos\frac{\pi}{2}$).

$\text{Area} = \int_0^{\pi/4}(\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi}(\sin x - \cos x)\,dx$

$= \left[\sin x + \cos x\right]_0^{\pi/4} + \left[-\cos x - \sin x\right]_{\pi/4}^{\pi}$

First integral: $\sin\frac{\pi}{4} + \cos\frac{\pi}{4} - (\sin 0 + \cos 0) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 1 = \sqrt{2} - 1$.

Second integral: $(-\cos\pi - \sin\pi) - (-\cos\frac{\pi}{4} - \sin\frac{\pi}{4}) = (1 - 0) - \!\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2}$.

$\text{Area} = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2}$

IT-2: Volume of Revolution with Substitution (with Algebra)

Question:

The region bounded by $y = \dfrac{1}{\sqrt{x + 1}}$, $x = 0$, $x = 3$And the $x$-axis is rotated $360\degree$ about the $x$-axis. Find the volume generated.

[Difficulty: hard. Combines volume of revolution with algebraic substitution and definite integral evaluation.]

Solution:

$V = \pi\int_0^3 \left(\frac{1}{\sqrt{x+1}}\right)^2 dx = \pi\int_0^3 \frac{1}{x + 1}\,dx$

$= \pi\Big[\ln(x + 1)\Big]_0^3 = \pi(\ln 4 - \ln 1) = \pi \ln 4 = 2\pi \ln 2$